0

Show that for any a in $\mathbb{R}$

$$\lim_{n\to ∞} \frac{a^n}{n!} = 0. $$ Hint: There exists a $n\in\mathbb{N}$ such that $n > |a|.$

I really do not know how to begin here with the proof and I would appreciate your help.

Jack D'Aurizio
  • 353,855
Polynor
  • 1
  • 1
    See this. – David Mitra Nov 28 '14 at 10:51
  • 1
    @DavidMitra: second question on this topic in last hour, and second answer using the series test to prove elementary convergence of a sequence in two days. – SBF Nov 28 '14 at 10:52
  • sry i dont have checked previously asked question but i thank you for your answer! – Polynor Nov 28 '14 at 10:56
  • @Ilya At least, no one's suggested to use Stirling :) – David Mitra Nov 28 '14 at 10:58
  • @DavidMitra: I'm pretty sure the only reason for that is that we just haven't waited enough yet :) – SBF Nov 28 '14 at 10:59
  • @Ilya: I should read the comments before I answer. My answer was too close to some of the others, but seeing as this is a duplicate to several questions, I should not have answered in the first place. – robjohn Nov 28 '14 at 11:12
  • @robjohn: by no means I was blaming you – SBF Nov 28 '14 at 11:34
  • @Ilya: I know... I was just noting that there were a lot of answers to a question which has been asked several times and I was one of them – robjohn Nov 28 '14 at 12:04

3 Answers3

1

Pick any such $N$ and pick $r$ such that $\dfrac{|a|}{N}<r<1$. For all $n>N$

$$\dfrac{|a|^n}{n!} = \underbrace{\left|\dfrac{a}{1}\dfrac{a}{2}\dfrac{a}{3}\cdots \dfrac{a}{N}\right|}_{:= \ C}\left(\dfrac{|a|}{N+1}\dfrac{|a|}{N+2}\dfrac{|a|}{N+3}\cdots \dfrac{|a|}{n}\right)< Cr^{n-N}$$

Now $r<1$. So?

David P
  • 12,320
1

Since: $$\prod_{n=1}^{N}\left(1+\frac{1}{n}\right)=N+1$$ we have: $$ n! = \frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}>e\cdot\left(\frac{n}{e}\right)^n \tag{1}$$ because for any $k\in\mathbb{N}^*$ we have $\left(1+\frac{1}{k}\right)^k<e$. A straightforward consequence of $(1)$ is that: $$\left|\frac{a^n}{n!}\right|<\frac{1}{e}\left(\frac{|a|e}{n}\right)^n,$$ hence, assuming $n\geq 2e|a|$, $$\left|\frac{a^n}{n!}\right| < \frac{1}{2^n}\to 0.$$

Jack D'Aurizio
  • 353,855
0

If $T_m=\dfrac{a^m}{m!},$

$$\lim_{n\to\infty}\frac{T_{n+1}}{T_n}=\lim_{n\to\infty}\frac a{n+1}=0$$ for finite $a$

So, $\sum T_r$ must converge $\implies\lim_{n\to\infty}T_n=0$

Alternatively for $N>|a|,$ $$\dfrac{a^m}{N!}<\frac aN$$

lab bhattacharjee
  • 274,582